day8

8:32 pm, 6 September 2025
My Jannik Sinner arc starts right now. the next two years of my life are jannik sinners jan 2018 to november 2019. It’s actually crazy, he was unranked at the start of jan 2018, playing completely average. well i mean not average really, like good. kind of like the level i am at right now. good but not the best in the world type stuff. by the end of the year, his skills were completely unrecognizable. The next year, he was defeating everybody and his ranking could not keep up with his results. and by the end of the next year, ie. november 2019, he was playing in the next gen finals. this is going to be my YC or startup thing. by the end of these two years, i’ll have my next gen torunament results. but right now tho, enjoy the journey you’re going on, this is so much fun, im literally living my dream bro, and the cool thing is the whole journey is documented here on my daily blogs ^_^ woohoo. yeah lets get to work then. ill work patiently and just trust the process,

it’s time to work and get ranked then.
9:16 pm
I’m doing Evan Chen’s probabilistic method handout A really cool problem is, if you take a subgraph of $K_{n,n}$ with at least $n^2-n+1$ edges, then you can always find a perfect matching. I couldn’t really think of a probabilistc proof, but i found a really cool proof by induction. Like if you let $F(n)=n^2-n+1$ , you can see that $F(n)-F(n-1)=2n-2$. So I mean let the two 'sets' be $A$ and $B$. In $A$ there must be a vertex with $n$ neighbours in $B$ (cos avg edges per vertex $=\frac{F(n)}{n} > n-1$. So I mean yeah, just choose any vertex in $B$ which does not have $n$ neighbours also (possible cos $=\frac{F(n)}{n} < n$ lmao) and like say these two vertices were $a,b$. Then consider $G-\{a,b\}$. We have removed at max $(n)+(n-1)-1=2n-2$ edges, hence this new graph $G'$ has at least $F(n-1)$ edges and hence it has a perfect matching, and since $a$ has n neighbours, we can append $b-a$ to the start of the path to get a new perfect matching!!! btw this was a perfect case of base case mattering in induction, cos if you notice, we just used the fact that $\Delta F = 2n-2$ , So I mean could $F(n)$ be $n^2-n$ also? The answer is nooooo, cos then the base case, $n=1$ is not true haha... And in general the bound is tight cos of $K_{n,n-1}$ and add an isolated vertex ^_^
12:08 am
well yuh, the probabilstic proof is pretty cool too, anyways, yeah i just feel so happy and like inspired now that ive decided to gte this, like fr i feel so happy. i watched thompson twins interview, i mean like theire vision board
is like kind of crazy, so inspirational man, like yuh we just gone follow our path and all our wildest dreams are gonna come true. anyways, yessir perfect, we back to math now. doing the classic ramsey example!
12:22 am
well the classic ramseey thing of showing $R(k,k)>2^{\frac{k}{2}}$ was ez… i remember reading about this earlier too, i think in noga alons book… hmm , i shoudl really go to sleep now, cos i gotta wake up tomorrow for kuku’s ioqm, but yeah, tmrw tho, what i think i should do is not stop at all, like overdo it, and the way to do that is like no internet, no breaks nothing, we just gonna work all day, and remain locked in woohoo… good night gng, get straight to it :D
btw tmrw is also kuku’s ioqm , i love you man , ioqm huh.. lol i remember i started math not knowing nothing, like nothing at all, and then i was 13th in India, you know you can be the best ever, so go be the best ever then man lesgo work bruh. all the way up.
5:54 pm, 7 September
Kuku’s ioqm over… now we’re back to work lesgo. not even gone hold you man, just go be the best there has ever been.
7:04 pm
Ok Sagnik had just called, we were looking at the pre-smmc paper at iitd (which i qualified obv), and there was this problem that said
$\binom{2p-1}{p-1} \equiv 1 (\mod p^3)$
My proof was write this as $(p+1)(p+2)..(p+p-1)=1.2.\cdots p-1$ and then expand the left hand side as a polynomial in $p$, you only need to expand till $p^2$ cos obv, and then the rest is very ez, like the linear term is $(p-1)!(\sum \frac{1}{i})$ which is known to be $0 \mod p^2$ in like MONT or wtv, and then the next term is the quadratic which is $\sum_{i \neq j} \frac{1}{ij}$ which you can again prove is $0 \mod p$ by looking at $(1+2+\cdots p-1)^2=1^2+2^2+\cdots (p-1)^2 + 2\sum_{i \neq j} ij$ and the first two terms are ezily divisible by $p$ woohooo!!!! apparently this is wolstenholme’s theorem
btw i said this to sagnik after:

and like yeah this is it, go have fun man. i promise, imma get this shit. like fr, life is so fucking good. jannik sinner arc right, go achieve all of your dreams every single dayyyyyyy!!!! locked in man woohooo.
9:12 pm
lol idk why kuku is so emotional, i love you kuku, you just gotta be happy cos your living your dream man, thats really it.
12:16 am, 8 september
ok man, i think i figured it out, the way to get this is to just work so hard, like work every single moment, so you dont even get a chance to think about nothing else right, yeah like get so lost in the work, nothing else matters… i told bunty mausi to bring Alon and Spencers probabilistic method book, life is so fucking good, you can actually just live yuor dream life every single day, how can you not lmaoooo
ok so in reading Evan’s handout, I came across this problem… this seems interesting and yk how i do shit, mony method right, so this is perfect, I wanna think about this completely independently so yuh…
The problem is:
Given a graph $G$, that is triangle free and has average degree $d$ , prove you can find large independent sets. (the bound in the handout is $0.01 * \frac{N}{d} \log (d)$ . I know Turan or something also gives you some bound, i dont know what it is at all, cos ive never read it. Also, I know this apparently has some connection to R(3,x) (ramsey number) that i’ll think about…
ok yuh thats the thing for right now.. but fr now that we’re just working hard, life is so fun, I can collab in my AI course with Daksh Kundu, and we can do cool AI shit together… I can collab in blockchain and crypto stuff with the blocsoc guys right, arnav panjla bhaiya, muskan kumari didi, sahitya shankar and more, I also told Dhruv Gaud and Sagnik Baidya that we’re gonna do olympiadish mocks every week together. so yeah life is fun.
btw i wrote everyone’s full names cos hopefully these names also mean something some day ^_^
uhh lets think about triangle free graphs now haha
12:45 am
I think I just redsicovered *some* theorem (I think Turan)

Ok, so just like consider the max independent set in any graph, call is MS_I and say it has size M. clearly, every other vertex in the graph must be adjacent to atleast some vertex in MS_I (else we contradict maximality cos you can just add that vertex.) So,
$N-M=V/MS_I=\bigcup_{v \in MS_I} N(v) \le \sum_{v \in MS_I} d(v) \le \Delta*M \implies \frac{N}{\Delta+1} \le M$
Ok yeah perfect, I didn’t even use the fact that graph is triangle free and we still got this, so in general, for any grpah G with $N$ vertices and max degree $\Delta$ , there is an independent set of size at least $\frac{N}{\Delta+1}$
pretty cool right. now let’s see…
ok wait taking a quick glance at the ajtai komlos szemeredi paper, this is not the turans theorem, but *a* turan theorem… hmm also wait, they prove the $\frac{N}{d+1}$ result, where $d$ is the average degree , which is obv much stronger than the max degree thing i proved… let’s think about how I can improve my result, cos this was just simple extremeal combo, like very simple, ive done this kind of thing before, in say Diestel, but the average degree thing is much much more interesting… ok i gotta think about that now, cos I have no idea how to convert max degree to average degree (sidenote, I do remember in diestel there was this problem that said that in a graph with min degree $\delta$ and girth atleast $g$, there is atleast $n_{0} (\delta,g)$ vertices, where $n_{0}(\delta,g)~delta^{\frac{g}{2}}$ (you can look at the exact thing in diestel chapter 1) but yeah the point is they had stated that this same result could also be proved for $d$ instead of $\delta$
(ok everywhere $\delta,d,\Delta$ refer to min, average, max degree)
and diestel had said alon had proved this using probabilistc method, this reminds me so much of that haha… probabilistic method is so cool yessirski lets work
listening to skyscraper-geographer ; the jannik sinner song haha…. we’re working man, lesgo, just keep having fun lol… sinner lost the first set btw (us open finals going on)
hmhmhmhmhm ahhh math is fun, just keep thinking lol
1:23 am
well the \frac{N}{d} result is stronger than the 0.01 thingy for $d ~< e^100$ sooo i mean just an observation haha… im very tired now and hence im going to sleep while watching the us open and maybe reading the ai book for a while idk lets see. good night ?
5:48 pm 8 September
It’s a monday, but i decided not to go to college w mummy today, and instead stay at home. in the morning when i woke up, did stupid shit, so i decided to sleep for an hour and reset the day, and get this shit started the right way. like really how can you not do your absolute best, its crazy.
Anyways, I was bored in the morning, so I was looking at Yufei Zhao’s extremal combo lectures on youtube, he gave this really cool probabilistc proof of Turan’s theorem, (lol im saying turan’s so many times but this time I’m reffering to, in a $K_{r+1}$ free graph, whats the max edges you can have?)
This is interesting to me , cos i was facin the same problem in this quesiton, which is how the fuck do i introduce randomness lmao. so there yufei did:
In a $n$ vertex, $K_{r+1}$ free graph, randomly order the vertices left to right. starting from left, let $X$ be the set that consists of all vertices $v$ such that all vertices to the left of $v$ are neighbours of $v$ ; or you can say all non neighbours of $v$ are to the right of $v$. clearly $X$ is a clique (inductively prove this)
Then you can compute $E[X]$ via linearty of expectation, cos the probability that $v$ is in $X$ is just $\frac{1}{n-d(v)}$
Then we note that this value must be less than $r$ cos I mean no $r+1$ clique right, so then just use Jensen and you’re done… fascinating stuff… lets think about my independent set problem now….
life is actually fucking perfect, you can never ever complain about anything. like any issues in your life are litrally so fucking trivial compared to what we’re trynna do, its just funny lol. also everything else, like girls, parties everything feels so stupid to look at when im locked in right , like you got to be an actual retard mony to not work your absolute hardest right now… lesgo. woohoo, my fun personality is back haha goofy kid and idgaf what you think cos im smarter than you and pretty soon ill make your mom my personal stripper #6, bitch. ^_^
6:00 pm
oh wait does this argument just work for the $\frac{n}{d+1}$ result???
Ok,so i think I just proved in any graph (not just triangle free), there is an indpendet set larger than $\frac{n}{d+1}$.
The proof is again, order the vertices of this grpah randomly, and put em in a list, left to right. Let $X$ be a set that contains of call vertices which are not connected to any vertex to it’s left, or you can say $v \in X$ if all neighours of $v$ are to it’s right. Clearly $X$ is an independet set, and clearly $P(v \in X)=\frac{1}{d(v)+1}$
So, $E[X]=\sum \frac{1}{d(v)+1} \ge \frac{n}{d+1}$ by Jensen so we're done!!!
This shit is actually overpowered wtf lmaoooo.
Now you gotta think about how to incorporate triangle free into randomness… so that i can prove the stronger result. given how similar these two proofs were, surely there is some relation between these two Turan’s theorems right?
8:19 pm
I just scootied, and while doing it i realised like yeah this is really my life and i can just do whatever the fuck i feel like and have fun right… so i did and jumped off 10 m in the swimming pool, and yuh, now we back at my table, day 8 is finna be legendary mane.
8:29 well yeah, both the turan’s theorem say approx the same thing, the $K_{r+1}$ free thing exists right, so then you can just take the complement of both graphs and you find out the new statement is equivalent to the $\frac{n}{d+1}$ independent set theorem. damn interetsing… now finally we can think about triangle free stuff haha. 8:39 hmm this is interesting to think about, but listen man… i saw the handout and like this shits so cool, it ends with LLL (laszo local lemma or some shit)… i remember when i had first heard of this, this type of stuff seemed completely beyond my reach right, now im like one of the smartest kids in india, and i can actually do all of this stuff man. this handout ends with imo 2014/6, i can actually do all of this, i can actually be asia #1 at smmc, i can actually do all this research level math, build cool ass projects and be a whiz kid billionaire right… just gotta enjoy the process, that being said, i am going to upload a daily blog every single day before sleeping now. thats a promise. every single day before sleeping i am going to upload my daily blog, so i guess day 60 will end at october 30th, and yeah thats it. by that time, i will have completed everything in the 60 day thingy, i promise every single day i am going to make legendary man, lets fucking go bruh. 9:25 this is a problem. the standard state im in is boredom, like i just dont know what to do… very tuff man, like i want to study, i just. idk man. but ive wasted like 5 years, i coulda been imo and ioi gold and jee air 1 or wtv by now, no way im going to not achieve all my dreams, i want thsi shit too much bruh, fuck this. im just going to work non stop, work will solve everything i think. so just litreaaly, idk man, very annoying, i need to get in that flow where i just wanna wake up and work all day, i mean like i wanna do that, but i just cant. lmao yk what, just stop being a pussy ass nigga man, this your life, you can do anything nigga, fuck you bitch. lets go non stop right here. 9:34 idgaf nigga. 12:34 am siddhant gadodia just called me and askde if i wanted to 1v1 him… yk this just reminded me yk what, tf you thinking about nigga, you just supposed to enjoy life right. so just have fun and just do whatever the fuck you feel like haha, literally just live your dream life and enjoy every single moment. this shit is literally histroric man, very soon you’ll look back on this time so fucking fondly if you just do what you ****actually want to**** 2:24 am um so i put on hikari by jishle (i rmbr i used to listen to this all the time during inmo) and like i couldnt get any ideas at all for the triangle free thingy, so i just went forward and skipped it for now. i guess this is really the only way to do stuff right , think about stuff , but obv you wont be able to prove everything you try, so sometimes, just move forward, read other stuff and come back to it later , or maybe read the solution or wtv right… but yeah i did these three practice problems,

all of them were pretty much immediate for me. bro this is sad, i havent done any geuinely tough, where like genuinely tough means at first i have no idea how to do it but then i push through and do it, like i havent done any problem like that in so long .. i think the last time i did was when i proved maj and inv are equidistributed, but ive gotta feel like that every day. i guess i see the path now tho, you sit down, and like put on hikari or wtv, and then you just enjoy the process and work , and like do something every single second… htink about difficult problems, read theory, do wtv you feel like, have fun man. letsgo bruh. i guess again one important thing is, do not take the small breaks at all pls, like where you go look at twitter or whatsapp or wtv, like really just work and like alex wang said, overdo it bruh… thats it, yk this what you want to do right… keep going then man. this handout is so fucking inspiring btw. 2:35 this is such a brilliant technique haha… you wanna prove you can always find an indpendent set of size $\frac{n}{2d}$ (yessss nigga i know i proved a stronger result above, but this just the technique demo)
so now, you select each vertex randomly with probability $p$ (we’ll fix $p$ later) so you selected $np$ vertices (exp. value)
and then if any two vertices in your selected set are joined by an edge, remove one of those vertices, so if there are $e$ edges, you remove at max $e$ vertices, so your set has atleast $np-e$ vertices.
now note expected value of e is just , there’s $\frac{nd}{2}$ edges, and each edge is included with proability $p^2$ ; so expected number of vertices you seltected is:
$np(1-\frac{dp}{2})$ then just choose $p=\frac{1}{d}$ and we’re done… this is honestly brilliant haha… apperantly this is called alteration… i mean this is the point mony, you gotta think deeply about problems but you also gotta learn cool new techniques right. this is fun bruh,
2:53
i do not undeerstand LLL at all right now. im going to sleep, we finna wake up and get on really good session in before leaving for delhi… just have fun and enjoy life bruh.
but really man, like one day, you gotta do the overdo it thing and see how it feels, cos actually if youre not overdoing it, then im not doing it al all, cos like im a kid whos either all in or all out, so if youre not studying all day, then i wont be able to study even a little bit. so im all ni today, i promsie nigga, lets go.
11:38 am 9 September
perfect start to the day, i woke up, aint even look at the clock, just brushed, got in the shower and like within 10 minutes, im at my table. one full study session right here right now, this starts the journey man, enjoy!
bruhhhh i looked at pranav sriram combo chaper 9, and i seriously still have suchhhh a long way to go haha… its ok tho, thats the point of this shit, i love this… lets go work, theres so much stuff i wanna do. im telling you, learn cool shit, and build a lot of cool shit like zuck, like things that give you notereity. and youll be smart and risk taking and bold and that implies youre a whiz kid billionaire ^_^
12:03 pm
oh my godddd bruh, damn wtf. looking at LLL, and like so much research done around it, this is crazy, theres so much stuff i wanna and gotta do, combinatorics, number theory, crytpography, information theory, coding theory, ai, ml everything man ,so like theres no way you can not work every single second of your life lmao. this is too much stuff i wanna do to waste a single second, this is so much fun ahhhh. but also, like keep in context that you are kind of a genius, like top 15 score in inmo, and like it shoulda been top 5, and thats like off so little work frfr, then elmo silver, like scoring more than most imo silver medallists, smmc last year, you scored more than like indian imo team members right, so like and this is off like your lazy ass work ethic, where i havent worked hard at all. so now that i have decided to work hard, and i dedicate every sinlge second of my life to getting better, and i work really hard, what are you going to become. thats the crazy and exciting part ;) but yeah for now bro, just keep studying ^_^ ahhhhh this is actully so exciting man, have fun, life is literally perfect, you have so manyyyy opportunities lol lesgo.
2:38 pm
lmao i messed up, took a break, but idgaf, we back at it… we gonna study for like 90-100 mins then go back to iit delhi :D
3:02 pm
I’m thinking about LLL, and like there’s a very standard hypergraph colouring problem , so I’m looking at what a hypergraph is and reading some notes ect… but thsi is fun…. btw i mean i was going to do olympiad math anyways, cos its fun and i get to get better, but like smmc is such a wonderful chance/ opportunity for me right , like this is probably the last time in my life i get to do olympiad math, i have always known i can actually be the best in the world if i just work hard, but yk i didnt… i regret it so much, this time i wanna have no regrets… smmc will be my jannik sinner getting ranked moment. i promise bruh, i promise man… go have soooooo much fun every single day man, lets fucking go bruh…you can do anything, lets go all the way up man… just have fun all day and achieve all your dreams, I am going to be top 1 in asia at smmc, i am going to drop out of iit dehi and be a fucking billlionaire. all the way up
3:15
ahhh yufei zhao’s site is so inspiring and exciting… but at some point, you just gotta sit down and think lmao. so yup just sit and think deeply now.
2:02 am 10 september
uhh im back in iit. while walking into aravali, i realised, this is it. ill be walking down this path through this whole journey every single day right. after smmc asia #1 , after math research problems, after cool projects, after YC startup, after ive dropped out, after im a billionaire. haha this is fun man, lets fucking go work